Home
Class 12
CHEMISTRY
Two liquids A and B form an ideal soluti...

Two liquids A and B form an ideal solution of 1 mole of A and x moles of B is 550 mm of Hg. If the vapour pressures of pure A and B are 400 mm of Hg nd 600 mm of Hg respectively. Then x is-

A

1

B

2

C

3

D

4

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step-by-step, we will use the information provided about the ideal solution and apply Dalton's Law of Partial Pressures. ### Step 1: Understanding the Given Information We have: - Moles of liquid A = 1 mole - Moles of liquid B = x moles - Total vapor pressure of the solution (P_total) = 550 mm of Hg - Vapor pressure of pure A (P°_A) = 400 mm of Hg - Vapor pressure of pure B (P°_B) = 600 mm of Hg ### Step 2: Calculate the Mole Fractions The mole fraction of A (X_A) and B (X_B) can be calculated as follows: - Mole fraction of A (X_A) = moles of A / total moles = 1 / (1 + x) - Mole fraction of B (X_B) = moles of B / total moles = x / (1 + x) ### Step 3: Apply Dalton's Law of Partial Pressures According to Dalton's Law, the total vapor pressure of the solution can be expressed as: \[ P_{total} = P°_A \cdot X_A + P°_B \cdot X_B \] Substituting the values we have: \[ 550 = 400 \cdot \left(\frac{1}{1+x}\right) + 600 \cdot \left(\frac{x}{1+x}\right) \] ### Step 4: Simplify the Equation Multiply both sides of the equation by (1 + x) to eliminate the denominator: \[ 550(1 + x) = 400 + 600x \] Expanding the left side gives: \[ 550 + 550x = 400 + 600x \] ### Step 5: Rearranging the Equation Now, rearranging the equation to isolate x: \[ 550 - 400 = 600x - 550x \] This simplifies to: \[ 150 = 50x \] ### Step 6: Solve for x Now, divide both sides by 50: \[ x = \frac{150}{50} = 3 \] ### Final Answer Thus, the value of x is 3. ---
Promotional Banner

Similar Questions

Explore conceptually related problems

Two liquids A and B form an ideal soluton. The vapour pressure of pure A and pure B are 66 mm Hg and 88 mm Hg , respectively. Calculate the composition of vapour A in the solution which is equilbrium and whose molar volume is 36% .

Two liquids A and B form an ideal soluton. The vapour pressure of pure A and pure B are 66 mm Hg and 88 mm Hg , respectively. Calculate the composition of vapour A in the solution which is equilbrium and whose molar volume is 36% .

Two liquids A and B form ideal solution. At 300 K , the vapour pressure of a solution containing 1 mole of A and 3 moles of B is 550 mm of Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. Determine the vapour pressure of a and B in their pure states.

Two liquids, A and B form an ideal solution. At the specified temperature, the vapour pressure of pure A is 20 mm Hg while that of pure B is 75 mm Hg . If the vapour over the mixture consists of 50 mol percent A , what is the mole percent A in the liquid?

An ideal solution is formed by mixing of 460 g. of ethanol with xg. of methanol. The total vapour pressure of the solution is 72 mm of Hg. The vapour pressure of pure ethanol and pure methanol are 48 and 80 mm of Hg respectively. Find the value of x. [Given: Atomic mass H = 1, C = 12, O = 16]

Two liquids A and B form an ideal solution. At 300 K , the vapour pressure of a solution containing 1 mol of A and 3 mol fo B is 550 mm Hg . At the same temperature, if 1 mol more of B is added to this solution, the vapour pressure of the solution increases by 10 mm Hg . Determine the vapour pressure of A and B in their pure states.

Benzene and naphthalene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of naphthalene.

Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300K are 50.71 mm Hg and 32.06mm Hg , respectively. Calculate the mole fraction of benzene in vapour phase if 80g of benzene is mixed with 100g of naphthalene.

The liquid A and B form ideal solutions. At 300 K, the vapour pressure of solution containing 1 mole of A and 3 mole of B is 550 mm Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg).

At 300 K , the vapour pressure of an ideal solution containing one mole of A and 3 mole of B is 550 mm of Hg . At the same temperature, if one mole of B is added to this solution, the vapour pressure of solution increases by 10 mm of Hg . Calculate the V.P. of A and B in their pure state.